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Proof step 5 from proof in-ecs-n (previous step)

n0 < # wlar, ∀ wlar, m. n0 ≤ # wlar → (m ∈ ecs(wlar, n0) ↔ (∃ n1. n1 < n0 ∧ wlar[n1].ec = m)) ⊦ m ∈ ecs(wlar, n0) ++ wlar[n0].ec ↔ (∃ n. ¬ n0 < n ∧ wlar[n].ec = m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
wlar[n0] = wl-entry(m0, stat), n0 < # wlar, ∀ wlar, m. n0 ≤ # wlar → (m ∈ ecs(wlar, n0) ↔ (∃ n1. n1 < n0 ∧ wlar[n1].ec = m)) ⊦ m ∈ ecs(wlar, n0) ++ m0 ↔ (∃ n. ¬ n0 < n ∧ wlar[n].ec = m)

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[n0], m0, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[n0], m0, stat